 #1
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 6
 Homework Statement:

Consider a very large flat surface located in the plane [itex]z = 0[/itex] that is uniformly loaded with a density [itex]\sigma =3\, \textrm{nC/m}^2[/itex]. Consider also a point charge [itex]q=100\, \textrm{pC}[/itex] located at point [itex](0,0,20)\, \textrm{cm}[/itex].
a) What force acts on the charge [itex]q[/itex]?
b) What is the flux of the electric field through a spherical surface of [itex]10\, \textrm{cm}[/itex]radius centered on the coordinate origin?
c) What force would act on a point charge [itex]Q=200\, \textrm{pC}[/itex] if we placed it at point [itex](0,5,20)\, \textrm{cm}[/itex]?
d) In the situation of the previous section, what would be the flux of the electric field through a spherical surface of [itex]10\, \textrm{cm}[/itex] radius centered on the charge [itex]Q[/itex]?
 Relevant Equations:

[tex] \vec{F}=\vec{E}\cdot q[/tex]
[tex] \phi =\oint \vec{E}d\vec{S}[/tex]
a)
[tex] \vec{F}=\vec{E}\cdot q [/tex]
[tex] \phi =\oint \vec{E}d\vec{S}=\oint \vec{E}d\vec{S}=\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \perp}+\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \parallel}=0+\oint EdS\cdot \underbrace{\cos 0}_1= E2S[/tex]
[tex] \dfrac{Q_{enc}}{\varepsilon_0}=\phi [/tex]
[tex]
\left.
\dfrac{Q_{enc}}{\varepsilon_0}=E2S \atop
\sigma =\dfrac{Q_{enc}}{S}
\right\} \quad \sigma S=E2S\varepsilon_0 \rightarrow E=\dfrac{\sigma}{2\varepsilon_0}=\dfrac{3\cdot 10^{9}}{2\cdot 8'85\cdot 10^{12}}=169'49\, \widehat{k}\, \textrm{N/C}
[/tex]
[tex] \vec{F}=\dfrac{\sigma}{2\varepsilon_0}\cdot q=\dfrac{3\cdot 10^{9}}{2\cdot 8'85\cdot 10^{12}}\cdot 1\cdot 10^{10}=1'69\cdot 10^{8}\, \textrm{N}\rightarrow \boxed{\vec{F}=17\widehat{k}\, \textrm{nN}} [/tex]
b)
[tex] \phi =\dfrac{Q_{enc}}{\varepsilon_0}=\dfrac{\pi \cdot R^2\sigma}{\varepsilon_0}=\dfrac{\pi \cdot 0'1^2\cdot 3\cdot 10^{9}}{8'85\cdot 10^{12}}=\boxed{10'65\, \textrm{Nm}^2/\textrm{C}} [/tex]
c) [tex] Q=200\, \textrm{pC}=2\cdot 10^{10}\, \textrm{C}\quad P_Q=(0,0'05,0'2)\, \textrm{m}\quad P_q=(0,0,0'2)\, \textrm{m} [/tex]
[tex] \vec{E}=169'49\, \widehat{k} [/tex]
[tex] \vec{F_E}=Q\cdot E=2\cdot 10^{10}\cdot 169'49\, \widehat{k}=3'39\cdot 10^{8}\, \textrm{C}\, \widehat{k} [/tex]
[tex] \vec{r_{Q\to q}}=(0,0'05,0'2)(0,0,0'2)=(0,0'05,0) [/tex]
[tex] \vec{F_q}=k\cdot \dfrac{Q\cdot q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{(2\cdot 10^{10})\cdot 1\cdot 10^{10}}{0'05^3}\cdot (0,0'05,0)=7'2\cdot 10^{8}\, \textrm{C}\, \widehat{j} [/tex]
[tex] \boxed{\vec{F}=(72\, \widehat{j}34\, \widehat{k})\, \textrm{nN}} [/tex]
d) [tex]
\left.
\vec{E}=169'49\, \widehat{k} \atop
\vec{E_q}=k\cdot \dfrac{q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{1\cdot 10^{10}}{0'05^3}(0,0'05,0)=360\, \widehat{j}
\right\} \vec{E_T}=(360\, \widehat{j}+169'49\, \widehat{k})\, \textrm{N/C}
[/tex]
How would section (d) of this question be calculated?
In section (b) I have applied Gauss law since I could calculate the internal charge of the spherical surface.
It is the one I've made so far. The problem I see is that I can't apply Gauss. But I also don't know the relationship between the direction of the surface differential and the direction of the electric field.
Thanks!
[tex] \vec{F}=\vec{E}\cdot q [/tex]
[tex] \phi =\oint \vec{E}d\vec{S}=\oint \vec{E}d\vec{S}=\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \perp}+\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \parallel}=0+\oint EdS\cdot \underbrace{\cos 0}_1= E2S[/tex]
[tex] \dfrac{Q_{enc}}{\varepsilon_0}=\phi [/tex]
[tex]
\left.
\dfrac{Q_{enc}}{\varepsilon_0}=E2S \atop
\sigma =\dfrac{Q_{enc}}{S}
\right\} \quad \sigma S=E2S\varepsilon_0 \rightarrow E=\dfrac{\sigma}{2\varepsilon_0}=\dfrac{3\cdot 10^{9}}{2\cdot 8'85\cdot 10^{12}}=169'49\, \widehat{k}\, \textrm{N/C}
[/tex]
[tex] \vec{F}=\dfrac{\sigma}{2\varepsilon_0}\cdot q=\dfrac{3\cdot 10^{9}}{2\cdot 8'85\cdot 10^{12}}\cdot 1\cdot 10^{10}=1'69\cdot 10^{8}\, \textrm{N}\rightarrow \boxed{\vec{F}=17\widehat{k}\, \textrm{nN}} [/tex]
b)
[tex] \phi =\dfrac{Q_{enc}}{\varepsilon_0}=\dfrac{\pi \cdot R^2\sigma}{\varepsilon_0}=\dfrac{\pi \cdot 0'1^2\cdot 3\cdot 10^{9}}{8'85\cdot 10^{12}}=\boxed{10'65\, \textrm{Nm}^2/\textrm{C}} [/tex]
c) [tex] Q=200\, \textrm{pC}=2\cdot 10^{10}\, \textrm{C}\quad P_Q=(0,0'05,0'2)\, \textrm{m}\quad P_q=(0,0,0'2)\, \textrm{m} [/tex]
[tex] \vec{E}=169'49\, \widehat{k} [/tex]
[tex] \vec{F_E}=Q\cdot E=2\cdot 10^{10}\cdot 169'49\, \widehat{k}=3'39\cdot 10^{8}\, \textrm{C}\, \widehat{k} [/tex]
[tex] \vec{r_{Q\to q}}=(0,0'05,0'2)(0,0,0'2)=(0,0'05,0) [/tex]
[tex] \vec{F_q}=k\cdot \dfrac{Q\cdot q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{(2\cdot 10^{10})\cdot 1\cdot 10^{10}}{0'05^3}\cdot (0,0'05,0)=7'2\cdot 10^{8}\, \textrm{C}\, \widehat{j} [/tex]
[tex] \boxed{\vec{F}=(72\, \widehat{j}34\, \widehat{k})\, \textrm{nN}} [/tex]
d) [tex]
\left.
\vec{E}=169'49\, \widehat{k} \atop
\vec{E_q}=k\cdot \dfrac{q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{1\cdot 10^{10}}{0'05^3}(0,0'05,0)=360\, \widehat{j}
\right\} \vec{E_T}=(360\, \widehat{j}+169'49\, \widehat{k})\, \textrm{N/C}
[/tex]
How would section (d) of this question be calculated?
In section (b) I have applied Gauss law since I could calculate the internal charge of the spherical surface.
It is the one I've made so far. The problem I see is that I can't apply Gauss. But I also don't know the relationship between the direction of the surface differential and the direction of the electric field.
Thanks!